Advertisements
Advertisements
Question
Without using the distance formula, show that the points A(4, −2), B(−4, 4) and C(10, 6) are the vertices of a right-angled triangle.
Solution
The given points are A(4, −2), B(−4, 4) and C(10, 6).
Slope of AB = `(4 + 2)/(-4 - 4)`
Slope of AB = `6/(-8)`
Slope of AB = `(-3)/4` ...(1)
Slope of BC = `(6 - 4)/(10 + 4)`
Slope of BC = `2/14`
Slope of BC = `1/7` ...(2)
Slope of AC = `(6 + 2)/(10 - 4)`
Slope of AC = `8/6`
Slope of AC = `4/3` ...(3)
From (1) and (3)
It can be seen that:
Slope of AB = `(-1)/"Slope of AC"`
Hence, AB ⊥ AC
Thus, the given points are the vertices of a right-angled triangle.
APPEARS IN
RELATED QUESTIONS
Find the slope of the line perpendicular to AB if : A = (0, −5) and B = (−2, 4)
Find the slope of the line which is perpendicular to `x - y/2 + 3 = 0`
Angles made by the line with the positive direction of X–axis is given. Find the slope of these line.
60°
Fill in the blank using correct alternative.
A line makes an angle of 30° with the positive direction of X– axis. So the slope of the line is ______.
Find the slope of a line, correct of two decimals, whose inclination is 50°
Find the slope of a line parallel to the given line x +3y = 7
Find the slope and the y-intercept of the following line 3x + y = 7
Find slope of a line passing through the points A(3, 1) and B(5, 3).
Find the slope of the line passing through the points A(6, -2) and B(–3, 4).
Show that points A(– 4, –7), B(–1, 2), C(8, 5) and D(5, – 4) are the vertices of a parallelogram ABCD
